CODEWITHSUNDEEP
c++templatesc++11metaprogramming
mkmostafa
mkmostafa

Reputation: 3171

template parameter with conversion operator

#include <iostream>

template <typename T1, typename T2> 
bool func(const T1& t, const T2& t2) {
    return t == t2;
}


class Base {
public:
    bool operator ==(const Base&) const { return true;}
    Base(int y) : x(y) {}
    operator int() {
        return x;
    }
    int x;
};

int main() {
    func<long, Base>(4L, Base(5)); // not ok
    func<long, long>(4L, Base(5)); //ok
}

Can somebody elaborate why the first version does not work? In otherwords why does the binary operator == in func not use the conversion operator int to convert the template parameter bound to Base into int?

Is there anyway to make version 1 work by only modifying the class Base?

Upvotes: 1

Views: 103

Answers (1)

Andrew
Andrew

Reputation: 5352

Your func accepts its parameters by const reference, but the operator int() defined in your Base class is a non-const member function. Mark it as a const member function, as shown below, and your code will compile:

#include <iostream>

template <typename T1, typename T2> 
bool func(const T1& t, const T2& t2) {
    return t == t2;
}


class Base {
public:
    bool operator ==(const Base&) const { return true;}
    Base(int y) : x(y) {}
    operator int() const {
        return x;
    }
    int x;
};

int main() {
    func<long, Base>(4L, Base(5)); // ok now!
    func<long, long>(4L, Base(5)); // ok
}

Live example

Upvotes: 3

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